Like most people, when I first encountered $n!$ in grade school, I graphed it, then connected the dots with a smooth curve and reasoned that there must be some meaning to $\left(\frac43\right)!$ — and, true to form, there was!

Gamma function (Image taken from Wikipedia.)

$$\displaystyle \Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,\rm dt \,.$$ Complex analysis means that integrals like this one make sense over $\mathbb{C}$, so, since this function is the same as $f(n) = n!$ on $\mathbb{N}$ (shifted by one), we call it an extension of the factorial. (Is it the only complex analytic function that is an extension? No, but with a few more restrictions, we can make $\Gamma(z)$ the "only" answer to the perfect interpolation question about $n!$.)

Naïvely, I look at the first few values of the partition number $p(n)$, and want to do the same thing. But, the partitions have no simple explicit formula; that's part of their mystery!

plot of the partition numbers p(n)

Emory math professor Ken Ono explains the state of things in this video. (Here is the relevant paper, as well.) There are some details that I still do not understand about the paper, but in the final steps before Bruinier and Ono give their explicit formula, they introduce a method that only makes sense when applied to a natural number, so we get a very cool formula... but not something that gives us a way of thinking about $p(z)$; it's still $p(n)$.

So, my question is this: is this a fool's errand? Can we say that there will never be a meaning for something like $p(2.6)$ or $p(3-\pi i)$? I think that this could be the case, because, if such an interpretation existed, then the problem of counting partitions would be much easier than it has proven to be thus far.

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    $\begingroup$ Perhaps you could say some more about the "few more restrictions"? I'd expect that the problem as you pose it is likely to be as underdetermined as the one for the factorial is without further restrictions. $\endgroup$
    – joriki
    Apr 21, 2011 at 16:34
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    $\begingroup$ It seems to me that Rademacher's Formula, mentioned in the first page of the paper you linked to, makes sense for complex values of $n$. However, it's not clear that this yields an interesting function. You should realize that the Gamma function is not "just" an interpolation of the factorial function; it is an extremely useful function in its own right. $\endgroup$
    – Alon Amit
    Apr 21, 2011 at 16:45
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    $\begingroup$ In Rademacher's formula you find $A_k(n) = \sum_{h=1}^k \exp(D-2\pi i n h/k)$ where $D$ is independent of $n$ (it does depend on $h$ and $k$). There's no reason why you can't plug in a complex value for $n$ here. $\endgroup$
    – Alon Amit
    Apr 21, 2011 at 18:02
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    $\begingroup$ I thought about this a lot this afternoon, it seems like theres a mysterious relationship between the partition function and gamma function. For one thing $1/\Gamma(z)$ is 0 on all negative integers and then takes on non-zero values (like the partition f) but also if you stick it blindly into the functional equation for the partition func $p(z) = \frac{1}{\Gamma(z+1)}$ obeys the identity $ p(z-1) + p(z-2) +... = p(z)*(EulerPhi(z)-1) $. Which is algebraically is VERY close to being equal for certain $z$ and moreover 1/EulerPhi anyways generates the coefficients of the partition function. $\endgroup$ Jan 13, 2021 at 23:38
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    $\begingroup$ The fact blindly applying the partition function functional equation to the reciprocal of the gamma function gives you multiplication by the generating function of the partition function cannot be some kind of "accidental" coincidence. $\endgroup$ Jan 13, 2021 at 23:42

1 Answer 1


(Not a full answer, but too long for a comment.)

One possible path to generalizing the partition function $p(n)$ to $p(\alpha)$ for complex $\alpha$ would be to consider the following series:

$$f(x)=1+\sum_{j=1}^\infty p(j) x^j=\frac1{(x;x)_{\infty}}=\sqrt[3]{\frac{2\sqrt[8]{x}}{\vartheta_1^{\prime}\left(0,\sqrt{x}\right)}}$$

where $(q;q)_{\infty}$ is a q-Pochhammer symbol and $\vartheta_1^{\prime}\left(0,q\right)$ is the derivative of the Jacobi theta function $\vartheta_1(z,q)$ with respect to $z$, evaluated at $z=0$. From this, we can define $p(n)$ as

$$p(n)=\frac1{n!}\left.\frac{\mathrm d^n}{\mathrm dx^n}f(x)\right|_{x=0}$$

and then consider if we can assign a meaning to

$$\frac1{\Gamma(\alpha+1)}\left.\frac{\mathrm d^\alpha}{\mathrm dx^\alpha}f(x)\right|_{x=0}$$

for arbitrary $\alpha$.

One way to go about this would be to consider Cauchy's differentiation formula

$$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\mathrm dz$$

for some appropriate closed contour $\gamma$ going anticlockwise, and then let

$$p(\alpha)=\frac1{2\pi i}\oint_\gamma \frac{f(z)}{z^{\alpha+1}}\mathrm dz$$

I'm a bit far away from my Mathematica computer to do a quick investigation for this, but turning this into a practical formula would involve making a proper choice of the integration contour $\gamma$, taking into account the fact that the function $z^{-\alpha-1}$ has a branch point at $z=0$, as well as $(q;q)_{\infty}$ not being defined for $|q| > 1$ (and having a lot of singularities on the boundary $|q| = 1$).

Hopefully somebody can carry through with this line of attack.

(added 4/30/2011)

I managed to borrow a machine with Mathematica, and decided to try experimenting with this construction. As mentioned, a suitable contour for the previously-mentioned integral must be an anticlockwise path about $z=0$, and should not cross the branch cut of $z^{-\alpha-1}$. With this, I decided to let $\gamma$ be a circle of radius $\rho$ going anticlockwise and starting from the negative real axis. I imposed the restriction $0 < \rho < 1$ due to the restricted domain of the q-Pochhammer symbol.

In short, I considered the integral

$$f(\alpha,\rho)=\frac1{2\pi\rho^\alpha}\int_0^{2\pi}\frac1{(-\rho\exp(it);-\rho\exp(it))_{\infty}(-\exp(it))^\alpha}\mathrm dt$$

for various values of $\rho\in(0,1)$.

In Mathematica:

pcont[a_?NumericQ, r_?NumericQ] := 
 NIntegrate[1/(QPochhammer[-r Exp[I t]] (-Exp[I t])^a), {t, 0, 2 Pi}, 
    AccuracyGoal -> Infinity, Method -> "Oscillatory", 
    WorkingPrecision -> Precision[{a, r}]]/(2 Pi r^a) /;
    Precision[{a, r}] < Infinity && 0 < r < 1

At integer values of $\alpha$, any choice of $\rho$ yielded the expected result $f(n,\rho)=p(n)$; the fun starts when noninteger values are considered:

f(\alpha,\rho) plots

The values of $f(\alpha,\rho)$ at the integers are indicated with red dots.

As can be seen, $f(\alpha,\rho)$ provides a one-parameter family of "analytic continuations" for the partition function $p(n)$; however, it might be the case that there exists a function $\rho(\alpha)$ such that $f(\alpha,\rho(\alpha))$ is the "natural" analytic continuation $p(\alpha)$. Alternatively, there might be a choice of the contour $\gamma$ that is better/more natural than a circle. I don't know any more than this, and it may be a while before I get back to this subject.

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    $\begingroup$ If the problem was to analytically continue the partition function $q(n)$, the function $f(x)$ that should be substituted into the Cauchy differentiation formula is $\frac1{(x;x^2)_\infty}$ $\endgroup$ Apr 25, 2011 at 3:59
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    $\begingroup$ What about a kind of limit $\rho \to 1$ ? It seems the graph gets less wobbly with higher values of $\rho$. $\endgroup$ Apr 30, 2011 at 9:26
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    $\begingroup$ @Raskolnikov: The problem is that the q-Pochhammer is what becomes unruly for values of $\rho$ near 1; have a look for instance at the complex plane plots. The function has a really nasty boundary... $\endgroup$ Apr 30, 2011 at 9:37
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    $\begingroup$ This is reminiscent of the Hardy-Ramanujan circle method. $\endgroup$ Apr 30, 2011 at 12:07
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    $\begingroup$ I suspect a thorough reading of both the Hardy/Ramanujan and Rademacher papers will afford more insight (especially the last one, since Rademacher also proceeded to derive his formula through Cauchy). $\endgroup$ Apr 30, 2011 at 12:07

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